1. Direction of translation

The line $L$ is $x-y=4 \implies y=x-4$ so its slope is $1$.

A line parallel to $L$ has direction vector proportional to $(1,1)$.

The unit vector along this direction is $\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right).$

Since the point is translated by $2\sqrt3$ units parallel to $L$, the displacement vector is $2\sqrt3\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)=\left(\sqrt6,\sqrt6\right)$ or in the opposite direction, $\left(-\sqrt6,-\sqrt6\right).$

So from $(2,1)$, the two possible translated points are $Q_1=(2+\sqrt6,1+\sqrt6),$ $Q_2=(2-\sqrt6,1-\sqrt6).$

2. Use the quadrant condition

We need $Q$ to lie in the third quadrant, so both coordinates must be negative.

For $Q_1$, both coordinates are positive, so it is not in the third quadrant.

For $Q_2$, $2-\sqrt6<0, \qquad 1-\sqrt6<0$ since $\sqrt6>2$. Thus, $Q=(2-\sqrt6,\;1-\sqrt6).$

3. Equation of line through $Q$ perpendicular to $L$

Since $L$ has slope $1$, any line perpendicular to $L$ has slope $-1$.

Hence its equation is of the form $x+y=c.$

Passing through $Q=(2-\sqrt6,1-\sqrt6)$, $c=(2-\sqrt6)+(1-\sqrt6)=3-2\sqrt6.$

Therefore the required line is $x+y=3-2\sqrt6.$

4. Match with options

This corresponds to Option C.